### Step Index Optical Fiber and Acceptance Angle#### Problem StatementConsider a step index optical fiber. Light enters from air at one end of the fiber at an angle θ with respect to the fiber axis. Show that there is a range of θ (called the "acceptance angle") outside which the light will no longer be able to propagate inside the fiber. Find the expression of the acceptance angle in terms of \( n_1 \) and \( n_2 \). Calculate the value for the case of \( n_1 = 1.5 \) and \( n_2 = 1.49 \).#### Diagram ExplanationThe accompanying diagram illustrates the cross-section of a step index optical fiber with the following components:- **Core (n1)**: The central region of the fiber with a refractive index of \( n_1 \).- **Cladding (n2)**: The outer layer surrounding the core with a slightly lower refractive index of \( n_2 \).- **Angle of Entry (θ)**: The angle at which light enters the fiber relative to the fiber axis.The diagram highlights the path of light as it enters the fiber core and is reflected internally, maintaining its propagation within the fiber as long as it is within the acceptance angle.#### Calculation of Acceptance Angle1. **Critical Angle**: Total internal reflection occurs at the core-cladding interface. \[ \sin(\theta_c) = \frac{n_2}{n_1} \]2. **Acceptance Angle**: Use Snell's law at the air-core interface. \[ \sin(θ_{accept}) = n_{core} \sin(\theta_c) \]3. **Expression in Terms of Refractive Indices**:\[ \sin(\theta_{accept}) = \sqrt{n_1^2 - n_2^2} \]4. **Substitute the Given Values**: \[ \begin{align*} n_1 &= 1.5 \\ n_2 &= 1.49 \\ \sin(\theta_{accept}) &= \sqrt{(1.5)^2 - (1.49)^2} \\ &= \sqrt{2.25 - 2.2201} \\ &= \sqrt{0.0299} \\ &= 0.173 \\ \theta_{accept} &= \

Given: The refractive index of the core is, n1 =1.5. The refractive index of the cladding is, n2…

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